Do Bosons Annihilate or Interact Differently?

[selfAdjoint]

Do boson anihilate?

This question was raised in my mind by a post on the Strings, Branes et al. board that had a link to a text describing gravitons anihilating and producing graviphotons. Aside from what you think of this model, what is the skinny on bosons anihilating?

Photons don't, AFAIK; do gluons? What about those massive weakons? Is it a wave function thing like the exclusion principle?

 

[lethe]

Originally posted by selfAdjoint
This question was raised in my mind by a post on the Strings, Branes et al. board that had a link to a text describing gravitons anihilating and producing graviphotons. Aside from what you think of this model, what is the skinny on bosons anihilating?

Photons don't, AFAIK; do gluons? What about those massive weakons? Is it a wave function thing like the exclusion principle?

its not clear what you mean by "do bosons annihilate". if you mean, "do bosons annihilate when they meet their antiparticles?" then the answer is yes.

of course, photons are their own antiparticles, so they can annihilite with other photons. that is, there is a vacuum bubble where two photons come out of the vacuum, and then annihilate.

but i don t know quite what to make of the question as it stands. a single particle never annihilates.

 

[jcsd]

selfAdjoint, think about it, one of the most well known anihilations is that of a low energy electron and a low energy positron which produces two photons, as anihilation is time reversable photon-photon anihilation must be possible:

$$e^-e^+ \longleftrightarrow \gamma\gamma$$

 

[MathematicalPhysicist]

Originally posted by lethe
its not clear what you mean by "do bosons annihilate". if you mean, "do bosons annihilate when they meet their antiparticles?" then the answer is yes.

of course, photons are their own antiparticles, so they can annihilite with other photons. that is, there is a vacuum bubble where two photons come out of the vacuum, and then annihilate.

but i don t know quite what to make of the question as it stands. a single particle never annihilates.

i have a question:
when electron and positron anihlate each other the product is energy now what happens when a photon anihlate another photon? (from the experience of mine when you point the flashlight to another flashlight no energy burst occur).

 

[mormonator_rm]

First of all, bosons do annihilate with their anti-particles. Photons with photons to produce electron-positron pairs (most commonly), gluons with gluons to produce quark-antiquark pairs, Z particles with Z particles to produce either lepton-antilepton pairs of quark-antiquark pairs, and W- with W+ to produce photons (generally, but sometimes Z particles). Bosons usually annihilate to produce matter that they would act upon. Hence, photons produce charged leptons readily, gluons produce colored quarks readily, and Z and W particles produce flavored quarks, leptons, and photons (as photons can have isospin) readily.

Photons annihilate to form matter, usually leptonic in the first family of leptons (i.e. electron-positron pairs).

I doubt that gravitons will annihilate to produce "graviphotons"; I think gravitons are probably analoguous to photons already (if they exist at all). It is possible, since we have bosons that decay into leptonic pairs and quark pairs already, that we need some boson that can decay into neutrino pairs to complete the model. The graviton could fit in this spot for one main reason; neutrinos, being nearly massless, could be easily produced from low energy graviton annihilation, and this could be another factor for the abundance of free neutrinos in nature.

 

[lethe]

Originally posted by mormonator_rm
It is possible, since we have bosons that decay into leptonic pairs and quark pairs already, that we need some boson that can decay into neutrino pairs to complete the model.

cannot the Z boson decay into neutrinos?

 

[mormonator_rm]

I would think so, since neutrinos do have isospin. It is constrained, however; in order to produce neutrinos, the decay/annihilation must also produce an equal number of photons. The branching ratio in the Particle Listings is so small that it is defined only by an upper bound. The branching fraction for neutrinos+photons over total width is < 3.1 x 10^-6, and the invariant mass of the photon-photon system emitted with the neutrino-antineutrino pair should be around 60 +/- 5 GeV. This means that at best it is rare, if it occurs at all. You can find these results in the Physical Review under Gauge and Higgs Bosons at http://pdg.lbl.gov/.

So, it appears that Z annihilation and decay to neutrino pairs is not a serious contributor to neutrino production. Hence my argument about gravitons still has some possibility of not being excluded easily (although it is very precarious at best). W-decay is a better producer of neutrinos, it only accounts for 30% of W-decays. But that is a leptonic decay, not an annihilation.

 

[meteor]

Given that the electron is the quantum of the electron field, and the photon the quantum of the electromagnetic field,there's such a thing like a positron field, being the positron the quantum? I'm not sure about this
I would appreciate a clarification about how the annihilation of two photons can yield an electron-positron pair. It suggests that the electromagnetic field and the electron field are connected somehow

 

[jcsd]

The Dirac field which describes electrons also describes positrons, they are essientally different states in the same field.

As mormonator indicated an electron-positron pair is not the only possible product of a photon-photon annihilation, I know that a baryon-antibaryon pair is possible as well, it's dependent on other factors like the photons' energies.

Wait for mormonator to give you a fuller answer tho'.

 

[meteor]

So there are eight different gluon fields, one for each gluon, or are the eight gluons different states of the same field?
The same question for the W⁺ and W^- bosons. Have they its own field or are different states of a single W field?

 

[jcsd]

The Dirac field is just about where my techincal knowledge of QFT starts and ends (even then I wouldn't say I have partciluarly stellar knowled of trhe Dirac field beyond the introductory chapter to QFT in my QM textbook)that's not to say I'm completely ignorant of other QFTs), but think about it there's only one weak field and there's only one strong field, so why should there be more fields when these fields are quantitzed?

As i siad before Mormonator has a much, much better handle on these things than me.

 

[mormonator_rm]

jcsd is correct on all counts so far. The positron acts as little more than a positively charged electron. You can think of them as coming from only one type of particle, but being able to carry a choice of two different (but opposite) electrical charges. Its actually a little more complicated than that, but that is really all you need to understand for now. W- and W+ are very similar to electrons in this way; they too can be treated as just oppositely charged renditions of the same particle. We have to throw in Z, though, and then the idea has to encompass the neutral member as well.

This same idea for W-, W+, and Z applies for the charged and neutral pions, and also for the proton and neutron in a sense. In the limit of isospin symmetry, these particles would manifest themselves as just one particle per set: for the pions, only the neutral pion would remain; for the proton and neutron, these would simply merge into one state, the nucleon; for the W-, W+, and Z, these would simply become W. However, this symmetry does not exist perfectly in nature, so we see a splitting of the degeneracy withtin these states.

Gluons are the same way; there is only one gluon, but you can give it anyone of eight color-charges. There aren't eight different kinds of gluons, just the same gluon carrying eight different possible color-charges.

Because of flavour and isospin assymetry, this same idea cannot be adequately extended to the different quark flavors or lepton families at this time. However, in certain limits it could still apply.

 

[meteor]

Is possible that this list of fields are the constituents of the universe?
- gravitational field
- Dirac field
- quark field : gives rise to all types of quarks
- gluon field
- neutrino field : gives rise to the six kind of neutrinos
- muon field : gives rise to the muon
- tau field : gives rise to the tau
- weak field : gives rise to the Z, W+ and W- bosons
- electromagnetic field

That is, can we say that the universe is composed of this 9 fields and its excitations, plus perhaps another field causing the expansion (quintessence), and perhaps also a Higgs field?

 

[mormonator_rm]

Remember to separate the neutrino fields, as each type is different. So, in essence, you end up with three massive lepton fields (electon/positron, muon/anti-muon, tau/anti-tau), three neutrino fields (e/anti-e, mu/anti-mu, tau/anti-tau), six quark fields (up/anti-up, down/anti-down, strange/anti-strange, charm/anti-charm, bottom/anti-bottom, top/anti-top), the weak field (W-, W+, Z), the gluon field, the electromagnetic field, and the gravitational field. The fields of massive leptons and their neutrinos are related, as also the fields of the weak and electromagnetic forces are related. Also the up and down quark fields are sufficiently degenerate to be considered one field (isospin). So that knocks it down to 11 fields.

From there we can reduce it even further. Each of the quark fields and gluon field (strong force) are corelated, as are each of the lepton fields and the unified electro-weak field. This brings us to only three fields. There is supposedly an anti-gravity type of field out there, but that probably is related to gravity as we know it already, and then there is the Higgs field which probably exists. That gives us four total.

The hope of many physicists is that we will come to understand the relationships between the electro-weak and strong fields, as well as confirming the role that the Higgs field appears to play in each of those interactions. Such a trail of discoveries would leave us with only two fields: the unified quantum-mechanical field of GUT, and the field of gravity. These last two, however, appear to be so diametrically opposed that some physicists doubt we will ever link them together.

If a TOE can be developed that successfully incorporates both GUT and gravity in theory and in experimentation, then we will understand that only one field operates in this universe, and that we just see different aspects of it in different situations. At this point, some people believe this is possible, and others do not. But most scientists in the realm of high-energy physics are actively engaged in trying to find out if this is truly the case.

 

[selfAdjoint]

I would like to return to the subject of photons. Is it really so that real photons annihilate? I have seen several papers online that boast about achieving two-photon interactions in high energy experiments and then note that the two photon interaction is not annihilation sensa strictu.

 

[lethe]

Originally posted by selfAdjoint
I would like to return to the subject of photons. Is it really so that real photons annihilate? I have seen several papers online that boast about achieving two-photon interactions in high energy experiments and then note that the two photon interaction is not annihilation sensa strictu.

photon-photon scattering is not a tree level process, perhaps that s what they mean.

but photon-photon annihilation is tree level.

 

[hey.like]

If the space time is sole only

As commonly concept, the time-space is one only. that is true ,it is hard to fact the fact. if there is a universe t-s and mass t-s and the later is unpermit to measure. and the event is always happen near some mass. perhaps this idea is a way.

 

[Bob3141592]

When an electron meets with a positron, does it have to annihilate? I was always under the impression that it does.

Further, it doesn't matter what kinetic energies the particles have. Momentum will be conserved by the angles the resultant photons fly off at. The photons will appear to fly off in opposite directions as seen by the frame of reference in which the original
electron-positron pair had equal kinetic energies. Is that right?

Now then, when two photons of exactly the same energy meet, do they have to annihilate? I don't see how this can be so, since photons intersect other photons all the time. And even if the two photons are of different energies, there is always a frame of reference where those two photons have the same energy. So that cannot restrict the annihilation.

So I'm confused, which is my typical state. If photons can annihilate, how can there be any of them left?

By the way (separate question): How can I spell check a reply before submitting it? My typing skills are poor, and spell check is an indispensable tool. But I don't see one here.

 

[Bob3141592]

Originally posted by lethe
photon-photon scattering is not a tree level process, perhaps that s what they mean.

but photon-photon annihilation is tree level.

What does "tree level" mean?

 

[jcsd]

Originally posted by Bob3141592
When an electron meets with a positron, does it have to annihilate? I was always under the impression that it does.

Further, it doesn't matter what kinetic energies the particles have. Momentum will be conserved by the angles the resultant photons fly off at. The photons will appear to fly off in opposite directions as seen by the frame of reference in which the original
electron-positron pair had equal kinetic energies. Is that right?

Now then, when two photons of exactly the same energy meet, do they have to annihilate? I don't see how this can be so, since photons intersect other photons all the time. And even if the two photons are of different energies, there is always a frame of reference where those two photons have the same energy. So that cannot restrict the annihilation.

So I'm confused, which is my typical state. If photons can annihilate, how can there be any of them left?

By the way (separate question): How can I spell check a reply before submitting it? My typing skills are poor, and spell check is an indispensable tool. But I don't see one here.

An electron colliding with a positron will anihilate, but no they don't (necessarily) have the same kinetic energy as for an electron and a postiron kinetic energy is a function of speed. The particles produced by anilhation are dpendent on the kinetic energies of the partilces being annihilated and photon pair production by electron-postiron anihilation is only the low energy case.

Four-momentum is the quantity that has to be conserved and energy and momentum are components of four-momentum and must both be conserved in all rest frames.

I don't kow a great deal about photon-photon anihilation, so in won't venur a guess at the exact mechanism.

 

[lethe]

Originally posted by Bob3141592
When an electron meets with a positron, does it have to annihilate? I was always under the impression that it does.

no, they can scatter off each other without annihilating (although on a quantum level, this is indistinguishable from an annihilation).

Further, it doesn't matter what kinetic energies the particles have. Momentum will be conserved by the angles the resultant photons fly off at. The photons will appear to fly off in opposite directions as seen by the frame of reference in which the original
electron-positron pair had equal kinetic energies. Is that right?

yeah, that s correct.

Now then, when two photons of exactly the same energy meet, do they have to annihilate? I don't see how this can be so, since photons intersect other photons all the time. And even if the two photons are of different energies, there is always a frame of reference where those two photons have the same energy. So that cannot restrict the annihilation.

electron-positron annihilation is always kinematically favored. photon-photon annihilation is only kinematically allowed if the photons have enough energy to produce some pair of charged particles. the lightest charged particle is the electron, with .5 MeV, so the photons must be quite high energy to exhibit photon-photon annihilation.

the photons coming out of your lightbulb cannot do it.

Originally posted by Bob3141592
What does "tree level" mean?

tree level is a level of interaction where only one virtual particle is exchanged: there are no loops in the feynman diagram

 

[Bob3141592]

Originally posted by lethe
electron-positron annihilation is always kinematically favored. photon-photon annihilation is only kinematically allowed if the photons have enough energy to produce some pair of charged particles. the lightest charged particle is the electron, with .5 MeV, so the photons must be quite high energy to exhibit photon-photon annihilation.

the photons coming out of your lightbulb cannot do it.

That all sounds perfectly reasonable, and I’d have been surprised if it were otherwise. But I’m still not convinced.

Let’s say I’m sitting on one side of a really big room, and in each corner of the opposite wall are two lamps. I see two rays, say each one averaging about 5,000 angstroms, converging on my eyes. I agree, as far as I know the photons about to collide have no where near enough energy to produce an electron and positron.
But that’s only as far as I know. Directly behind me and halfway from the moon, is a neutrino traveling at just a shade less than the speed of light itself. In the reference frame of the neutrino, those two photons are sufficiently blueshifted that they do seem to have enough energy to annihilate into an electron and positron pair.
Who’s to say? Does the fact that the photons do not annihilate indicate that my frame of reference is to be preferred over the neutrino’s? I thought stuff like that wasn’t allowed.
So I’m still confused. I’m pretty sure about what the answer ought to be, but stuff like this makes me unsure of the justification. And I’m not trying to be difficult, really. I’m just trying to understand.

 

[selfAdjoint]

The photons are of low energy in the frame of the room. The only frame in which they are blue-shifted (in your thought experiment) is that of the neutrino. If the neutrino comes to interact with them, without slowing down, that energy will come into the interaction (from the perspective of the room, it is the neutrino that has brought the high energy). But in the room they are just visual light photons, without enough energy to make a pair of paricles.

 

[lethe]

Originally posted by Bob3141592
But that’s only as far as I know. Directly behind me and halfway from the moon, is a neutrino traveling at just a shade less than the speed of light itself. In the reference frame of the neutrino, those two photons are sufficiently blueshifted that they do seem to have enough energy to annihilate into an electron and positron pair.
Who’s to say?

they don t have enough center of mass energy, which is the kind of energy i meant when i said that the photons need to have more energy than the rest mass of two electrons. if the neutrino does not participate in the reaction, then i should not include it when calculating the center of mass energy.

the outcome of the reaction is two electrons at rest, in the center of mass frame of the two electrons. so it is in that frame that the photons have to have more the 1 MeV of energy.

nothing particular special about this reference frame, it s just easy to do calculations here, so this is where my quoted number came from.

 

[Bob3141592]

Originally posted by selfAdjoint
The photons are of low energy in the frame of the room. The only frame in which they are blue-shifted (in your thought experiment) is that of the neutrino. If the neutrino comes to interact with them, without slowing down, that energy will come into the interaction (from the perspective of the room, it is the neutrino that has brought the high energy). But in the room they are just visual light photons, without enough energy to make a pair of paricles.

and

Originally posted by lethe
they don t have enough center of mass energy, which is the kind of energy i meant when i said that the photons need to have more energy than the rest mass of two electrons. if the neutrino does not participate in the reaction, then i should not include it when calculating the center of mass energy.

Thanks. That makes sense. The existence of a reference frame in and of itself doesn't affect what reactions are possible because it's not part of that reaction. Would it be fair to say this is because the neutrino is outside the light cone of the photons, and so isn't a part of their universe (that's an awkward wording--is there a better way)?

I've always found the ways of photons impossible to understand. Even though I find the answers to my question clear, it'll take me some time to think about it and put it together. Maybe it'll answer other questions I have, but more likely it'll open up other questions and leave me more confused. It seems my understanding of physics is more chaotic than asymptotic! Bet I'm not alone in that feeling, eh?

In any case, I'm glad I found this forum. I appreciate the time everyone puts into this area, and hope to participate more in the future.

 

[lethe]

Originally posted by Bob3141592
Would it be fair to say this is because the neutrino is outside the light cone of the photons, and so isn't a part of their universe (that's an awkward wording--is there a better way)?

no, i don t think that is accurate. if the two interacting particles meet the neutrino in the room, then the neutrino is definitely inside their light cones.

if the two particles were not photons, then the neutrino might interact with one or both of them, but then we re not talking about photon-photon annihilation, we re talking about something else.

also, photons cannot interact with neutrinos, since they are neutral.